resistance

Last updated 2023.03.02 Edit Source

• Physics
• Electricity

resistors
$$R=\frac{V}{I}\to I=\frac{V}{R}\to V=I\cdot R$$ ^476c14

electric power

Resistance of graphite
$d=0.7\text{mm}$
$l=7\text{cm}=0.07\text{m}$
$R=40\Omega$
$\rho_{\text{graphite}}=?$
$$R=\rho \cdot \frac{l}{A}$$
$[\rho]=\Omega m$
$\rho=\frac{\Omega mm^{2}}{m}$

$r=\frac{d}{2}=0.35$
$A=r^{2}\pi=$

Electric boiler
$P=2500\text{ W}$
$I=?$
$V=120\text{ l}$
$T_{\text{in}}=12\text{°C} \to T_{\text{out}}=60\text{°C}$
$c=4.2 \frac{\text{kJ}}{\text{kg°C}}$
$t=?$

$V=230\text{ V}$
$I=\frac{P}{V}=\frac{2500}{230}=10.9\text{ A}$
$$\Delta T=60-12=48\text{°C}$$
$W=\frac{120 \cdot 48}{4.2}=1370.43\text{ kWh}$

Solution
$$Q_{\text{taken by water}}=c \cdot m \cdot \Delta T=Q_{\text{delivered by boiler }}=P \cdot \Delta t$$
$4.2 \cdot 120(60-12)=2.5 \cdot \Delta t$
$t=9677\text{ s}=2.688\text{ h}$